What is the solution for x(dx)/(dt)+bt=0 ?

The solution for x(dx)/(dt)+bt=0 is x=sqrt(-t^2b+c_{1)},x=-sqrt(-t^2b+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x(dx)/(dt)+bt=0

x \frac{d x}{d t} + b t = 0

x = - t^{2} b + c_{1}, x = - - t^{2} b + c_{1}

Solution steps

x \frac{d x}{d t} + b t = 0

Solve separable ODE:

x = - t^{2} b + c_{1}, x = - - t^{2} b + c_{1}

x = - t^{2} b + c_{1}, x = - - t^{2} b + c_{1}

\frac{\partial}{\partial x} (2 y cos^{2} (x))

\int_{0}^{6} 4 x - 2 x^{2} d x

\int x (3 - x^{2})^{2} d x

x \to 3 lim (\frac{1}{x - 3} + 2)

4 t \frac{d y}{d t} + y = t^{2}

What is the solution for x(dx)/(dt)+bt=0 ?
The solution for x(dx)/(dt)+bt=0 is x=sqrt(-t^2b+c_{1)},x=-sqrt(-t^2b+c_{1)}